Find x-intercepts and vertex by symmetry and graphing

Example) Graph the given function and find a) the vertex b)the x-intercepts
F(x)=2x^2-4x+2

[A] a)(-1,8) b)1,-1

[B] a)(1,8) b)-4+2(5^3),-4-2(5^3)

[C] a)(-1,4) b)no x-intercepts

[D] a)(1,0) b)1

Grapmatica
Put the formula into graphmatica using the formula y=2x^2-4x+2. You should come up with a picture that looks like the one below.
graph.JPGThe vertex is the highest or lowest point on the parabola and the vertex for this graph is (1,0) and there is only one x-intercept and that is where that parabola goes through the x axis and that point is 1.
So the vertex is (1,0) and the x-intercept is 1. Therefore the correct answer is D.


Graphing Calculator
Putting formula in calculator
You can also use your calculator for this question. You would use the same formula. You would push the y= button and get to a list. Put the formula above in to Y1, Press graph, it will show you the graph and you can adjust your settings so that you can see the whole graph.

Finding vertex
To find the vertex you would push 2nd Trace, and whether the parabola opens up or down would depend on whether you chose minimum or maximum. In this case you would choose minimum. The screen will tell you to go left bound and then right bound of where you think the vertex is. It will then tell you the vertex. (1,0)

Finding x-intercepts
To find the x-intercepts you would go back into the 2nd Trace list and pick zero. The screen will once again tell you left bound and right bound of where you think it is. It will then tell you that the x-intercept is 1.
Once again the vertex is (1,0) and the x-intercept is 1.

You can use either a graphing calculator or graphmatica to come up with the same solutions it will just depend on which one you like better